# Understand the rank of $\begin{bmatrix} A&b\\ b^{*}&0 \end{bmatrix}$

Let $$A \in M_n(C)$$ and $$b$$ be a column vector of n complex complements. Denote $$\widetilde A = \begin{bmatrix} A&b\\ b^{*}&0 \end{bmatrix}$$ If $$rank(\widetilde A)=rank(A)$$, which of the following is true?

(a) $$Ax = b$$ has infinitely many solutions.

(b) $$Ax = b$$ has a unique solution.

(c) $$\widetilde Ax = 0$$ has only solution x = 0.

(d) $$\widetilde Ax = 0$$ has nonzero solutions.

Zhang, Fuzhen. Linear Algebra

I am assuming that $$b^*$$ is the conjugate transpose. Could you help me construct an example $$A$$ and $$\widetilde A$$ such as $$rank(\widetilde A)=rank(A)$$ ? I have hard time doing that without assuming that $$b=0$$.

Here is one example: $$\tilde A = \begin{pmatrix} -1 & 0 & -1 \\ 0 & 1 & 1 \\ -1 & 1& 0 \end{pmatrix}$$
Hint: If both matrices have the same rank (column rank = row rank), then $$b$$ lies in the column space of $$A$$ and so there is a linear combination of the columns of $$A$$ which gives $$b$$.
• Could you please construct a simple example $A$ , $b$ and $\widetilde A$? That's where I am struggling. Commented Aug 18, 2019 at 12:07
• For a trivial example, take $A=I, or$b=0$. Commented Aug 18, 2019 at 12:10 • Is it ever possible for b not be$0$? Commented Aug 18, 2019 at 12:12 • By the rank condition, yes its possible. Commented Aug 18, 2019 at 12:16 • @Wuestenfux: That condition is not sufficient, because the same linear combination might not work for the last row of$\tilde A\$. Commented Aug 18, 2019 at 13:02